F - The kth great number(优先队列)

F - The kth great number

Xiao Ming and Xiao Bao are playing a simple Numbers game. In a round Xiao Ming can choose to write down a number, or ask Xiao Bao what the kth great number is. Because the number written by Xiao Ming is too much, Xiao Bao is feeling
giddy. Now, try to help Xiao Bao.

InputThere are several test cases. For each test case, the first line of input contains two positive integer n, k. Then n lines follow. If Xiao Ming choose to write down a number, there will be an " I" followed by a number that Xiao Ming will write
down. If Xiao Ming choose to ask Xiao Bao, there will be a "Q", then you need to output the kth great number. 

OutputThe output consists of one integer representing the largest number of islands that all lie on one line. 

Sample Input

8 3
I 1
I 2
I 3
Q
I 5
Q
I 4
Q

Sample Output

1
2
3

Hint

Xiao  Ming  won't  ask  Xiao  Bao  the  kth  great  number  when  the  number  of  the  written number is smaller than k. (1=<k<=n<=1000000).

本题主要的意思就是说 根据输入的以上的数据要找到第k大的数

解题思路： 卡死宝宝了作为一个渣渣生存真的真的不容易超级不容易 刚开始用sort()结果愉快的发现真的不行  好吧只能用学长昨天讲的优先队列了  哎就是优先队列  最小堆  小的在前面

然后 把超过k个数并且已经排好队了把那些小的给剔除出去

C++语言代码：

#include<iostream>
#include<vector>
#include<stdio.h>
#include<queue>
using namespace std;
int main()
{
   int n,k,i,digit;
   char s[2];
   while(scanf("%d%d",&n,&k)==2)
   {
   	priority_queue<int ,vector<int>,greater<int> >q;    
   	for(i=1;i<=n;i++)
   	{
   		scanf("%s",s);
   		if(s[0]=='I')
   		{
   			scanf("%d",&digit);
   			q.push(digit);
   	        	if(q.size()>k)
   		       q.pop();
                }
   		else 
   		{
   			printf("%d\n",q.top());
   		}
   	}
   
   } 
   return 0;
}